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shawnkat
Hey there!

Normally, we would simplify the top using the distributive property which states that:

a(b + c) = ab + ac

However, we have a special situation here which makes the problem that much easier. Our first step is to take the denominator and put it in the form of a product of binomials. We know that we can use (x + 2)(x + 2) because when we use foil to solve this, we get x^2 + 2x + 2x + 4 = x^2 + 4x + 4

Therefore, we have:

[tex]4(x+2)/(x+2)(x+2)[/tex]

Now, we can cancel one pair of (x+2) because (x+2)/(x+2) = 1

That gives us just that one four over the x + 2:

[tex]4/x+2[/tex]

Hope this helps!

Astute
Hey there!

So, let's start out with figuring out the first part of this equation.

[tex] \left[\begin{array}{ccc}\boxed{\boxed{4(x+2}}\end{array}\right] [/tex]

Step #1

Simplify [tex] \frac{x+2}{y2} [/tex]

[tex]((4* \frac{(x+2)}{x^2} +4x) \ +4[/tex]

Step #2

[tex]( \frac{4 \ * \ (x+2)}{x^2} )+ \ \ 4x)+4[/tex]

Step #3

We would have to rewrite this whole fraction using [tex]\boxed{(x^2)}[/tex] as the denominator.

[tex]\boxed{\boxed{(4x= \frac{4x}{1} = \frac{4x+x^2}{x^2} )}}[/tex]

Step #4

We would have to combine the numerators.

[tex] \frac{4* \ (x+2) \ +4x \ * \ x^2}{x^2} [/tex] = [tex] \frac{4x^3 \ + \ 4x \ + \ 8}{x^2} [/tex]

By finishing what is above, it would look like [tex]\swarrow \swarrow \swarrow \swarrow [/tex]

                                      [tex] (\frac{4x^3+4x+8}{x^2} )[/tex]

Step #5

And now that we have come to know this from above, this leaves us with . . .

[tex]4(x+2)/(x+2)(x+2)[/tex]

(Final answer)

[tex] \left[\begin{array}{ccc}\boxed{\boxed{ \frac{4}{x}+2}} \end{array}\right] [/tex]

Hope this helps you!
~Jurgen

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